Two students have the same mean maths mark: $6{,}5$. One always gets similar marks ($6, 7, 6, 7$); the other alternates between top marks and fails ($3, 10, 3, 10$). The mean cannot tell these two very different situations apart: you need a measure of how far the data lie from their mean.
Before you start, make sure you know: Frequency tables (3rd ESO)
What dispersion measures
A first simple idea is the range (maximum minus minimum), but it only looks at the two extremes and ignores all the other data. To use all of the data, you measure the distance from each value to the mean, $x_i - \bar{x}$, and then average these distances.
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Two students have the following marks in mathematics: Student A: $6, 7, 6, 8, 7$ Student B: $4, 9, 5, 10, 6$ Without doing complex calculations, which of the two…
Two students have the following marks in mathematics:
Student A: $6, 7, 6, 8, 7$
Student B: $4, 9, 5, 10, 6$
Without doing complex calculations, which of the two students do you think has a larger standard deviation? Confirm your answer by calculating the standard deviation of both.
With data grouped into intervals, you use the class midpoints $x_i$ and their frequencies: $\sigma^2\approx\sum f_i(x_i-\bar x)^2/N$. This is an estimate of the original variance, because we do not know each individual value; it should not be confused with the exact variance of a table of individual values with frequencies.
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Calculate the variance and the standard deviation of the following data set, corresponding to the ages of a group of five friends: $15, 16, 15, 17, 16$.
Calculate the variance and the standard deviation of the following data set, corresponding to the ages of a group of five friends: $15, 16, 15, 17, 16$.
In a statistical study with $N = 40$ data values, it is known that the arithmetic mean is $\bar{x} = 5,2$ and that the sum of the squares of all data values is…
In a statistical study with $N = 40$ data values, it is known that the arithmetic mean is $\bar{x} = 5,2$ and that the sum of the squares of all data values is $\sum x_i^2 = 1250$. Using the shortcut formula, calculate the variance and the standard deviation.
The following table shows the time, in minutes, taken by employees of a factory to reach their workplace: $[0, 10)$: $12$ employees $[10, 20)$: $18$ employees…
The following table shows the time, in minutes, taken by employees of a factory to reach their workplace:
$[0, 10)$: $12$ employees
$[10, 20)$: $18$ employees
$[20, 30)$: $14$ employees
$[30, 40)$: $6$ employees
Calculate the variance and the standard deviation of the travel time using the class marks.
If you know the mean and variance of two groups separately, you can find the variance of the whole group without going back over all the data one by one, using the fact that $\sum x_i^2 = N(\sigma^2 + \bar{x}^2)$ in each group and adding.
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The standard deviation of the heights of a group of trees in a nursery is $12\text{ cm}$. If it is known that in one year all the trees will have grown by exactly…
The standard deviation of the heights of a group of trees in a nursery is $12\text{ cm}$. If it is known that in one year all the trees will have grown by exactly $15\text{ cm}$, what will the new standard deviation of their heights be? Justify your answer using the properties of measures of dispersion.
In a tech company, the variance of the monthly salaries of its employees is $40\,000\text{ €}^2$. If management decides to increase all salaries by $5\%$ (that is,…
In a tech company, the variance of the monthly salaries of its employees is $40\,000\text{ €}^2$. If management decides to increase all salaries by $5\%$ (that is, multiply them by $1,05$), what will the new variance of the salaries be?
A language academy has two B2 level groups. The morning group has $20$ students with a mean mark of $7$ and a variance of $1,5$. The afternoon group has $30$ students…
A language academy has two B2 level groups. The morning group has $20$ students with a mean mark of $7$ and a variance of $1,5$. The afternoon group has $30$ students with the same mean mark of $7$, but with a variance of $2,5$. Calculate the overall variance of all $50$ B2 level students combined.
Which is more spread out: the heights of a basketball team ($\sigma = 8\,\text{cm}$) or those of a gymnastics team ($\sigma = 7\,\text{cm}$)? They cannot be compared directly if the means are very different ($1{,}95\,\text{m}$ compared with $1{,}65\,\text{m}$): you need to make the comparison relative.
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In a 4th ESO class, the mean mark in a history exam was $6,5$ marks and the coefficient of variation is $20\%$. From these data, calculate the variance of the class's…
In a 4th ESO class, the mean mark in a history exam was $6,5$ marks and the coefficient of variation is $20\%$. From these data, calculate the variance of the class's marks.
In a basketball team, the mean height of the players is $1,95\text{ m}$ with a standard deviation of $0,08\text{ m}$. On the other hand, in a rhythmic gymnastics team,…
In a basketball team, the mean height of the players is $1,95\text{ m}$ with a standard deviation of $0,08\text{ m}$. On the other hand, in a rhythmic gymnastics team, the mean height is $1,65\text{ m}$ with a standard deviation of $0,07\text{ m}$. Which team has more dispersed heights in relative terms? Justify your answer numerically.
Back to the start: With marks that have a mean of $6{,}5$, the consistent student ($6, 7, 6, 7$) has a standard deviation of only $\sigma = 0{,}5$, whereas the erratic student ($3, 10, 3, 10$) reaches $\sigma = 3{,}5$.
Summary
Let's practice! 10 medium exercises in this section